Given a 2D array A[][2] of size N (1 ≤ N ≤ 103), where A[i][0] and A[i][1] denotes the length and breadth of rectangle i respectively.
Two rectangle i and j where (i < j) are similar if the ratio of their length and breadth is equal
A[i][0] / A[i][1] = A[j][0] / A[j][1]
The task is to count the pair of rectangles that are nearly similar.
Examples:
Input : A[][2] = {{4, 8}, {15, 30}, {3, 6}, {10, 20}}
Output: 6
Explanation: Pairs of similar rectangles are (0, 1), {0, 2), (0, 3), (1, 2), (1, 3), (2, 3). For every rectangle, ratio of length : breadth is 1 : 2Input : A[][2] = {{2, 3}, {4, 5}, {7, 8}}
Output: 0
Explanation: No pair of similar rectangles exists.
Approach: Follow the steps to solve the problem
- Traverse the array.
- For every pair such that (i < j), check whether the rectangles are similar or not by checking if the condition A[i][0] / A[i][1] = A[j][0] / A[j][1] is satisfied or not.
- If found to be true, increment the count.
- Finally, print the count obtained.
Below is the implementation of the above approach:
C++
// C++ Program for the above approach#include <iostream>using namespace std;// Function to calculate the count// of similar rectanglesint getCount(int rows, int columns, int A[][2]){ int res = 0; for (int i = 0; i < rows; i++) { for (int j = i + 1; j < rows; j++) { if (A[i][0] * 1LL * A[j][1] == A[i][1] * 1LL * A[j][0]) { res++; } } } return res;}// Driver Codeint main(){ // Input int A[][2] = { { 4, 8 }, { 10, 20 }, { 15, 30 }, { 3, 6 } }; int columns = 2; int rows = sizeof(A) / sizeof(A[0]); cout << getCount(rows, columns, A); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{ // Function to calculate the count // of similar rectangles static int getCount(int rows, int columns, int[][] A) { int res = 0; for(int i = 0; i < rows; i++) { for(int j = i + 1; j < rows; j++) { if (A[i][0] * A[j][1] == A[i][1] * A[j][0]) { res++; } } } return res; } // Driver Codepublic static void main(String[] args){ // Input int[][] A = { { 4, 8 }, { 10, 20 }, { 15, 30 }, { 3, 6 } }; int columns = 2; int rows = 4; System.out.print(getCount(rows, columns, A)); }}// This code is contributed by code_hunt. |
Python3
# Python3 program for the above approach# Function to calculate the count# of similar rectanglesdef getCount(rows, columns, A): res = 0 for i in range(rows): for j in range(i + 1, rows, 1): if (A[i][0] * A[j][1] == A[i][1] * A[j][0]): res += 1 return res# Driver Codeif __name__ == '__main__': # Input A = [ [ 4, 8 ], [ 10, 20 ], [ 15, 30 ], [ 3, 6 ] ] columns = 2 rows = len(A) print(getCount(rows, columns, A))# This code is contributed by SURENDRA_GANGWAR |
C#
// C# program for the above approachusing System;class GFG{// Function to calculate the count // of similar rectangles static int getCount(int rows, int columns, int[,] A) { int res = 0; for(int i = 0; i < rows; i++) { for(int j = i + 1; j < rows; j++) { if (A[i, 0] * A[j, 1] == A[i, 1] * A[j, 0]) { res++; } } } return res; } // Driver codestatic void Main(){ // Input int[,] A = { { 4, 8 }, { 10, 20 }, { 15, 30 }, { 3, 6 } }; int columns = 2; int rows = 4; Console.Write(getCount(rows, columns, A)); }}// This code is contributed by divyesh072019 |
6
Time Complexity: O(N2)
Auxiliary Space: O(1)
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